将分配转换为字典

Question

I have hard typed lots of assignments:

definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
.
.
.

Intend to transform them to dict avoiding repeating typing:

{'definition': ['basename', 'dirname', 'supports_unicode_filenames'],
 'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'] ...}

I tried to encapsulate them in class.

class OsPath:
    definition = ['basename', 'dirname', 'supports_unicode_filenames']
    condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']

Working on console

In [125]: dt = dict(vars(OsPath))
In [127]: {i:dt[i] for i in dt if not i.startswith('__')}
Out[127]:
{'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
 'definition': ['basename', 'dirname', 'supports_unicode_filenames']}

How to do it in shortcut?

Answer 1

You can do:

definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']
d = {'definition': definition, 'condition': condition }

Note that you should avoid using the variable name dict , which is a builtin object.

But to be honest, what you're trying to do is not good practice if you're doing this at scale. You should rethink the design of your object.

Answer 2

Could this be what you want:

definition = ['basename', 'dirname', 'supports_unicode_filenames']
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount']

result = {i: globals()[i] for i in ['definition', 'condition']}

Answer 3

Use SimpleNamespace

from types import SimpleNamespace
sn = SimpleNamespace(
definition = ['basename', 'dirname', 'supports_unicode_filenames'],
condition = ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
)

It outputs:

In [44]: sn
Out[44]: namespace(condition=['isabs', 'isdir', 'isfile', 'islink', 'ismount'], definition=['basename', 'dirname', 'supports_unicode_filenames'])

In [45]: vars(sn)
Out[45]:
{'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount'],
 'definition': ['basename', 'dirname', 'supports_unicode_filenames']}

In [17]: dict(definition = ['basename', 'dirname', 'supports_unicode_filenames'],condition = ['isabs', 'isdir'
    ...: , 'isfile', 'islink', 'ismount'])
Out[17]: 
{'definition': ['basename', 'dirname', 'supports_unicode_filenames'],
 'condition': ['isabs', 'isdir', 'isfile', 'islink', 'ismount']}